Branch: Mechanical Engineering Year: 1st

Branch: Mechanical EngineeringYear: 1st

Subject: Engineering GraphicsBy Bhupesh Kumar,

Lecturer,

Chapter 3Scales

➢

FOR FULL SIZE SCALE

R.F.=1 OR ( 1:1 )

MEANS DRAWING

& OBJECT ARE OF

SAME SIZE.

Other RFs are described

as

1:10, 1:100,

1:1000, 1:1,00,000

SCALES

DIMENSIONS OF LARGE OBJECTS MUST BE REDUCED TO ACCOMMODATE

ON STANDARD SIZE DRAWING SHEET. THIS REDUCTION CREATES A SCALE

OF THAT REDUCTION RATIO, WHICH IS GENERALLY A FRACTION..SUCH A SCALE IS CALLED REDUCING SCALE

AND

THAT RATIO IS CALLED REPRESENTATIVE FACTOR.

SIMILARLY IN CASE OF TINY OBJECTS DIMENSIONS MUST BE INCREASED

FOR ABOVE PURPOSE. HENCE THIS SCALE IS CALLED ENLARGING SCALE.

HERE THE RATIO CALLED REPRESENTATIVE FACTOR IS MORE THAN UNITY.

REPRESENTATIVE FACTOR (R.F.) =

=

=

=

A

USE FOLLOWING FORMULAS FOR THE CALCULATIONS IN THIS TOPIC.

B LENGTH OF SCALE = R.F. MAX. LENGTH TO BE MEASURED.X

DIMENSION OF DRAWING

DIMENSION OF OBJECT

LENGTH OF DRAWING

ACTUAL LENGTH

AREA OF DRAWING

ACTUAL AREA

VOLUME AS PER DRWG.

ACTUAL VOLUME

V

V3

1. PLAIN SCALES ( FOR DIMENSIONS UP TO SINGLE DECIMAL)

2. DIAGONAL SCALES ( FOR DIMENSIONS UP TO TWO DECIMALS)

3. VERNIER SCALES ( FOR DIMENSIONS UP TO TWO DECIMALS)

4. COMPARATIVE SCALES ( FOR COMPARING TWO DIFFERENT UNITS)

5. SCALE OF CORDS ( FOR MEASURING/CONSTRUCTING ANGLES)

TYPES OF SCALES:

= 10 HECTOMETRES

= 10 DECAMETRES

= 10 METRES

= 10 DECIMETRES

= 10 CENTIMETRES

= 10 MILIMETRES

1 KILOMETRE

1 HECTOMETRE

1 DECAMETRE

1 METRE

1 DECIMETRE

1 CENTIMETRE

BE FRIENDLY WITH THESE UNITS.

0 1 2 3 4 510

PLAIN SCALE:- This type of scale represents two units or a unit and it’s sub-division.

METERS

DECIMETERSR.F. = 1/100

4 M 6 DM

PLANE SCALE SHOWING METERS AND DECIMETERS.

PROBLEM NO.1:- Draw a scale 1 cm = 1m to read decimeters, to measure maximum distance of 6 m.

Show on it a distance of 4 m and 6 dm.

CONSTRUCTION:-

a) Calculate R.F.=

R.F.= 1cm/ 1m = 1/100

Length of scale = R.F. X max. distance

= 1/100 X 600 cm

= 6 cms

b) Draw a line 6 cm long and divide it in 6 equal parts. Each part will represent larger division unit.

c) Sub divide the first part which will represent second unit or fraction of first unit.

d) Place ( 0 ) at the end of first unit. Number the units on right side of Zero and subdivisions

on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a look of scale.

e) After construction of scale mention it’s RF and name of scale as shown.

f) Show the distance 4 m 6 dm on it as shown.

DIMENSION OF DRAWING

DIMENSION OF OBJECTPLAIN SCALE

Problem No 2: In a map a 36 km distance is shown by a line 45 cms long. Calculate the R.F. and construct

a plain scale to read kilometers and hectometers, for max. 12 km. Show a distance of 8.3 km on it.

CONSTRUCTION:-

a) Calculate R.F.

R.F.= 45 cm/ 36 km = 45/ 36 . 1000 . 100 = 1/ 80,000

Length of scale = R.F. max. distance

= 1/ 80000 12 km

= 15 cm

b) Draw a line 15 cm long and divide it in 12 equal parts. Each part will represent larger division unit.

c) Sub divide the first part which will represent second unit or fraction of first unit.


on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a look of scale.

e) After construction of scale mention it’s RF and name of scale as shown.

f) Show the distance 8.3 km on it as shown.

KILOMETERSHECTOMETERS

8KM 3HM

R.F. = 1/80,000

PLANE SCALE SHOWING KILOMETERS AND HECTOMETERS

0 1 2 3 4 5 6 7 8 9 10 1110 5

PLAIN SCALE

PROBLEM NO.3:- The distance between two stations is 210 km. A passenger train covers this distance

in 7 hours. Construct a plain scale to measure time up to a single minute. RF is 1/200,000 Indicate the distance

traveled by train in 29 minutes.

CONSTRUCTION:-

a) 210 km in 7 hours. Means speed of the train is 30 km per hour ( 60 minutes)

Length of scale = R.F. max. distance per hour

= 1/ 2,00,000 30km

= 15 cm

b) 15 cm length will represent 30 km and 1 hour i.e. 60 minutes.

Draw a line 15 cm long and divide it in 6 equal parts. Each part will represent 5 km and 10 minutes.

c) Sub divide the first part in 10 equal parts,which will represent second unit or fraction of first unit.

Each smaller part will represent distance traveled in one minute.


on left-hand side of Zero. Take height of scale 5 to 10 mm for getting a proper look of scale.

e) Show km on upper side and time in minutes on lower side of the scale as shown.

After construction of scale mention it’s RF and name of scale as shown.

f) Show the distance traveled in 29 minutes, which is 14.5 km, on it as shown.

0 10 20 30 40 5010 MINUTESMIN

R.F. = 1/100

PLANE SCALE SHOWING METERS AND DECIMETERS.

KMKM 0 5 10 15 20 255 2.5

DISTANCE TRAVELED IN 29 MINUTES.

14.5 KM

PLAIN SCALE

We have seen that the plain scales give only two dimensions, such

as a unit and it’s subunit or it’s fraction.

1

2

3

4

5

6

7

8

9

10X

Y

Z

The principle of construction of a diagonal scale is as follows.

Let the XY in figure be a subunit.

From Y draw a perpendicular YZ to a suitable height.

Join XZ. Divide YZ in to 10 equal parts.

Draw parallel lines to XY from all these divisions

and number them as shown.

From geometry we know that similar triangles have

their like sides proportional.

Consider two similar triangles XYZ and 7’ 7Z,

we have 7Z / YZ = 7’7 / XY (each part being one unit)

Means 7’ 7 = 7 / 10. x X Y = 0.7 XY

:.

Similarly

1’ – 1 = 0.1 XY

2’ – 2 = 0.2 XY

Thus, it is very clear that, the sides of small triangles,

which are parallel to divided lines, become progressively

shorter in length by 0.1 XY.

The solved examples ON NEXT PAGES will

make the principles of diagonal scales clear.

The diagonal scales give us three successive dimensions

that is a unit, a subunit and a subdivision of a subunit.

DIAGONAL

SCALE

R.F. = 1 / 40,00,000

DIAGONAL SCALE SHOWING KILOMETERS.

0 100 200 300 400 500100 50

109876543210

KMKM

KM

569 km

459 km

336 km

222 km

PROBLEM NO. 4 : The distance between Delhi and Agra is 200 km.

In a railway map it is represented by a line 5 cm long. Find it’s R.F.

Draw a diagonal scale to show single km. And maximum 600 km.

Indicate on it following distances. 1) 222 km 2) 336 km 3) 459 km 4) 569 km

SOLUTION STEPS: RF = 5 cm / 200 km = 1 / 40, 00, 000

Length of scale = 1 / 40, 00, 000 X 600 X 105 = 15 cm

Draw a line 15 cm long. It will represent 600 km.Divide it in six equal parts.( each will represent 100 km.)

Divide first division in ten equal parts. Each will represent 10 km.Draw a line upward from left end and

mark 10 parts on it of any distance. Name those parts 0 to 10 as shown.Join 9th sub-division of horizontal scale

with 10th division of the vertical divisions. Then draw parallel lines to this line from remaining sub divisions and

complete diagonal scale.

DIAGONAL

SCALE

PROBLEM NO.5: A rectangular plot of land measuring 1.28 hectors is represented on a map by a similar rectangle

of 8 sq. cm. Calculate RF of the scale. Draw a diagonal scale to read single meter. Show a distance of 438 m on it.

Draw a line 15 cm long.

It will represent 600 m.Divide it in six equal parts.

( each will represent 100 m.)

Divide first division in ten equal parts.Each will

represent 10 m.

Draw a line upward from left end and

mark 10 parts on it of any distance.

Name those parts 0 to 10 as shown.Join 9th sub-division

of horizontal scale with 10th division of the vertical divisions.

Then draw parallel lines to this line from remaining sub divisions

and complete diagonal scale.

DIAGONAL

SCALE

SOLUTION :

1 hector = 10, 000 sq. meters

1.28 hectors = 1.28 X 10, 000 sq. meters

= 1.28 X 104 X 104 sq. cm

8 sq. cm area on map represents

= 1.28 X 104 X 104 sq. cm on land

1 cm sq. on map represents

= 1.28 X 10 4 X 104 / 8 sq cm on land

1 cm on map represent

= 1.28 X 10 4 X 104 / 8 cm

= 4, 000 cm

1 cm on drawing represent 4, 000 cm, Means RF = 1 / 4000

Assuming length of scale 15 cm, it will represent 600 m.

0 100 200 300 400 500100 50

109876543210

M

M

M

438 meters

R.F. = 1 / 4000

DIAGONAL SCALE SHOWING METERS.engineering108.com

109876543210

CENTIMETRES

MM

CM

R.F. = 1 / 2.5

DIAGONAL SCALE SHOWING CENTIMETERS.

0 5 10 155 4 3 2 1

PROBLEM NO.6:. Draw a diagonal scale of R.F. 1: 2.5, showing centimeters

and millimeters and long enough to measure up to 20 centimeters.

SOLUTION STEPS:

R.F. = 1 / 2.5

Length of scale = 1 / 2.5 X 20 cm.

= 8 cm.

1.Draw a line 8 cm long and divide it in to 4 equal parts.

(Each part will represent a length of 5 cm.)

2.Divide the first part into 5 equal divisions.

(Each will show 1 cm.)

3.At the left hand end of the line, draw a vertical line and

on it step-off 10 equal divisions of any length.

4.Complete the scale as explained in previous problems.

Show the distance 13.4 cm on it.

13 .4 CM

DIAGONAL

SCALE

COMPARATIVE SCALES: These are the Scales having same R.F.

but graduated to read different units.

These scales may be Plain scales or Diagonal scales

and may be constructed separately or one above the other.

EXAMPLE NO. 7 :

A distance of 40 miles is represented by a line

8 cm long. Construct a plain scale to read 80 miles.

Also construct a comparative scale to read kilometers

upto 120 km ( 1 m = 1.609 km )

SOLUTION STEPS:

Scale of Miles:

40 miles are represented = 8 cm

: 80 miles = 16 cm

R.F. = 8 / 40 X 1609 X 1000 X 100

= 1 / 8, 04, 500

CONSTRUCTION:

Take a line 16 cm long and divide it into 8 parts. Each will represent 10 miles.

Subdivide the first part and each sub-division will measure single mile.

Scale of Km:

Length of scale

= 1 / 8,04,500 X 120 X 1000 X 100

= 14. 90 cm

CONSTRUCTION:

On the top line of the scale of miles cut off a distance of 14.90 cm and divide

it into 12 equal parts. Each part will represent 10 km.

Subdivide the first part into 10 equal parts. Each subdivision will show single km.

10 100 20 305 50 60 70 MILES40

10 0 10 20 30 40 50 60 70 80 90 100 110 KM

5

R.F. = 1 / 804500

COMPARATIVE SCALE SHOWING MILES AND KILOMETERS

COMPARATIVE SCALE:

EXAMPLE NO. 8 :

A motor car is running at a speed of 60 kph.

On a scale of RF = 1 / 4,00,000 show the distance

traveled by car in 47 minutes.

SOLUTION STEPS:

Scale of km.

length of scale = RF X 60 km

= 1 / 4,00,000 X 60 X 105

= 15 cm.

CONSTRUCTION:

Draw a line 15 cm long and divide it in 6 equal parts.

( each part will represent 10 km.)

Subdivide 1st part in `0 equal subdivisions.

( each will represent 1 km.)

Time Scale:

Same 15 cm line will represent 60 minutes.

Construct the scale similar to distance scale.

It will show minimum 1 minute & max. 60min.

10 100 20 305 50 KM40

10 100 20 305 50 MINUTES40

MIN.

KM

47 MINUTES

47 KM

R.F. = 1 / 4,00,000COMPARATIVE SCALE SHOWING MINUTES AND KILOMETERS

EXAMPLE NO. 9 :

A car is traveling at a speed of 60 km per hour. A 4 cm long line represents the distance traveled by the car in two hours.

Construct a suitable comparative scale up to 10 hours. The scale should be able to read the distance traveled in one minute.

Show the time required to cover 476 km and also distance in 4 hours and 24 minutes.

COMPARATIVE

SCALE:

SOLUTION:

4 cm line represents distance in two hours , means for 10 hours scale, 20 cm long line is required, as length

of scale.This length of scale will also represent 600 kms. ( as it is a distance traveled in 10 hours)

CONSTRUCTION:

Distance Scale ( km)Draw a line 20 cm long. Divide it in TEN equal parts.( Each will show 60 km)

Sub-divide 1st part in SIX subdivisions.( Each will represent 10 km)

At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length.

And complete the diagonal scale to read minimum ONE km.

Time scale:Draw a line 20 cm long. Divide it in TEN equal parts.( Each will show 1 hour) Sub-divide 1st part in SIX subdivisions.( Each will

represent 10 minutes) At the left hand end of the line, draw a vertical line and on it step-off 10 equal divisions of any length.

And complete the diagonal scale to read minimum ONE minute.

10

5

0

kM

kM 060 60 120 180 240 300 360 420 480 540

060 1 2 3 4 5 6 7 8 9

HOURS

MIN.

10

5

0

KILOMETERSDISTANCE SCALE TO MEASURE MIN 1 KM

TIME SCALE TO MEASURE MIN 1 MINUTE.

4 hrs 24 min. ( 264 kms )

476 kms ( 7 hrs 56 min.)

Figure to the right shows a part of a plain scale in

which length A-O represents 10 cm. If we divide A-O

into ten equal parts, each will be of 1 cm. Now it would

not be easy to divide each of these parts into ten equal

divisions to get measurements in millimeters.

Now if we take a length BO equal to 10 + 1 = 11 such equal

parts, thus representing 11 cm, and divide it into ten equal

divisions, each of these divisions will represent 11 / 10 – 1.1

cm.

The difference between one part of AO and one division of

BO will be equal 1.1 – 1.0 = 0.1 cm or 1 mm.

This difference is called Least Count of the scale.

Minimum this distance can be measured by this scale.

The upper scale BO is the Vernier. The

combination of plain scale and the Vernier is

Vernier scale.

Vernier Scales:These scales, like diagonal scales , are used to read to a very small unit with great accuracy.

It consists of two parts – a primary scale and a vernier. The primary scale is a plain scale fully

divided into minor divisions.

As it would be difficult to sub-divide the minor divisions in ordinary way, it is done with the help of the vernier.

The graduations on vernier are derived from those on the primary scale.

9.9 7.7 5.5 3.3 1.1

9 8 7 6 5 4 3 2 1 0A

0B

Example 10:

Draw a vernier scale of RF = 1 / 25 to read centimeters upto

4 meters and on it, show lengths 2.39 m and 0.91 m

.9 .8 .7 .6 .5 .4 .3 .2 .1

.99 .77 .55 .33 .11 01.1

0 1 2 31.0

SOLUTION:

Length of scale = RF X max. Distance

= 1 / 25 X 4 X 100

= 16 cm

CONSTRUCTION: ( Main scale)


Divide it in 4 equal parts.

( each will represent meter )

Sub-divide each part in 10 equal parts.

( each will represent decimeter )

Name those properly.

CONSTRUCTION: ( vernier)

Take 11 parts of Dm length and divide it in 10 equal parts.

Each will show 0.11 m or 1.1 dm or 11 cm and construct a rectangle

Covering these parts of vernier.

TO MEASURE GIVEN LENGTHS:

(1) For 2.39 m : Subtract 0.99 from 2.39 i.e. 2.39 - .99 = 1.4 m

The distance between 0.99 ( left of Zero) and 1.4 (right of Zero) is 2.39 m

(2) For 0.91 m : Subtract 0.11 from 0.91 i.e. 0.91 – 0.11 =0.80 m

The distance between 0.11 and 0.80 (both left side of Zero) is 0.91 m

1.4

2.39 m

0.91 m

METERSMETERS

Vernier Scale

Example 11: A map of size 500cm X 50cm wide represents an area of 6250 sq.Kms.

Construct a vernier scaleto measure kilometers, hectometers and decameters

and long enough to measure upto 7 km. Indicate on it a) 5.33 km b) 59 decameters.Vernier Scale

SOLUTION:

RF =

=

= 2 / 105

Length of

scale = RF X max. Distance

= 2 / 105 X 7 kms

= 14 cm

AREA OF DRAWING

ACTUAL AREAV

500 X 50 cm sq.

6250 km sq.V

CONSTRUCTION: ( vernier)

Take 11 parts of hectometer part length

and divide it in 10 equal parts.

Each will show 1.1 hm m or 11 dm and

Covering in a rectangle complete scale.

CONSTRUCTION: ( Main scale)


Divide it in 7 equal parts.

( each will represent km )

Sub-divide each part in 10 equal parts.

( each will represent hectometer )

Name those properly.

KILOMETERSHECTOMETERS

0 1 2 310 4 5 6

90 70 50 30 10

99 77 55 33 11Decameters

TO MEASURE GIVEN LENGTHS:

a) For 5.33 km :

Subtract 0.33 from 5.33

i.e. 5.33 - 0.33 = 5.00

The distance between 33 dm

( left of Zero) and

5.00 (right of Zero) is 5.33 k m

(b) For 59 dm :

Subtract 0.99 from 0.59

i.e. 0.59 – 0.99 = - 0.4 km

( - ve sign means left of Zero)

The distance between 99 dm and

- .4 km is 59 dm

(both left side of Zero)

5.33 km59 dm

100

200

300

400

500

600

700800 900

00

0 10 20 4030 7050 60 9080

SCALE OF CORDS

OA

CONSTRUCTION:

1. DRAW SECTOR OF A CIRCLE OF 900 WITH ‘OA’ RADIUS.( ‘OA’ ANY CONVINIENT DISTANCE )

2. DIVIDE THIS ANGLE IN NINE EQUAL PARTS OF 10 0 EACH.

3. NAME AS SHOWN FROM END ‘A’ UPWARDS.

4. FROM ‘A’ AS CENTER, WITH CORDS OF EACH ANGLE AS RADIUS

DRAW ARCS DOWNWARDS UP TO ‘AO’ LINE OR IT’S EXTENSION

AND FORM A SCALE WITH PROPER LABELING AS SHOWN.

AS CORD LENGTHS ARE USED TO MEASURE & CONSTRUCT

DIFERENT ANGLES IT IS CALLED SCALE OF CORDS.engineering108.com

100

200

300

400

500

600700

800 900

00

0 10 20 4030 7050 60 9080

OA

OA

B

O1 A1

B1

x

z

y

PROBLEM 12: Construct any triangle and measure it’s angles by using scale of cords.

CONSTRUCTION:

First prepare Scale of Cords for the problem.

Then construct a triangle of given sides. ( You are supposed to measure angles x, y and z)

To measure angle at x:Take O-A distance in compass from cords scale and mark it on lower side of triangle

as shown from corner x. Name O & A as shown. Then O as center, O-A radius

draw an arc upto upper adjacent side.Name the point B.

Take A-B cord in compass and place on scale of cords from Zero.

It will give value of angle at x

To measure angle at y:Repeat same process from O1. Draw arc with radius O1A1.

Place Cord A1B1 on scale and get angle at y.

To measure angle at z:Subtract the SUM of these two angles from 1800 to get angle at z.

SCALE OF CORDS

300550

Angle at z = 180 – ( 55 + 30 ) = 950

100

200

300

400

500

600700

800 900

00

0 10 20 4030 7050 60 9080

OA

Construct 250 and 1150 angles with a horizontal line , by using scale of cords.

CONSTRUCTION:

First prepare Scale of Cords for the problem.

Then Draw a horizontal line. Mark point O on it.

To construct 250 angle at O.Take O-A distance in compass from cords scale and mark it on on the line drawn, from O

Name O & A as shown. Then O as center, O-A radius draw an arc upward..

Take cord length of 250 angle from scale of cords in compass and

from A cut the arc at point B.Join B with O. The angle AOB is thus 250

To construct 1150 angle at O.

This scale can measure or construct angles upto 900 only directly. Hence Subtract 1150 from 1800.We get 750 angle ,

which can be constructed with this scale.

Extend previous arc of OA radius and taking cord length of 750 in compass cut this arc

at B1 with A as center. Join B1 with O. Now angle AOB1 is 750 and angle COB1 is 1150.

B1

750

1150

B

250

A O

OC

A

To construct 250 angle at O.To construct 1150 angle at O.

SCALE OF CORDS

Chapter 10

Screw Threads and Threaded Fasteners

External (male)

thread

Internal (female)

thread

THREAD TERMINOLOGY

A thread cut on the outside of

a cylindrical body.

A thread cut on the inside of

a cylindrical body.

Internal thread

External thread

Right-hand

thread

Thread that will assemble when

turned clockwise.

Left-hand

thread

Thread that will assemble when

turned counter-clockwise.

THREAD TERMINOLOGY

Turnbuckle use RH and LH thread at

each end to double displacement.

Crest

Root

Thread angle

THREAD TERMINOLOGY

The peak edge of a thread.

The bottom of the thread cut into

a cylindrical body.

The angle between threads faces.

Internal ThreadExternal Thread

CrestRoot

Thread angle

CrestRoot

Major diameterThe largest diameter on

an internal or external thread.

Minor diameterThe smallest diameter on

an internal or external thread.


Min

or

dia

.

Ma

jor

dia

.

THREAD TERMINOLOGY

Min

or

dia

.

Ma

jor

dia

.

PitchThe distance between crests of

threads.

LeadThe distance a screw will advance

when turned 360o.

THREAD TERMINOLOGY


PitchPitch

Form is the profile shape of the

thread.Thread Form

THREAD TERMINOLOGY

Example :

“knuckle thread form”

EXTERNAL THREAD CUTTING

Tools Operation

Threading Die

Die stock

Twist drill

Tools

Tap

Tap wrench

Operation

INTERNAL THREAD CUTTING

COMPARISON OF THREAD CUTTING

External Thread Internal Thread

Major Dia.

Minor Dia.

Thread Drawing

1. Detailed representation

2. Schematic representation

3. Simplified representation

THREAD REPRESENTATION

External thread Internal thread

DETAILED REPRESENTATION

60oPitch

Use slanting lines to represent crest and root.

Roots and crest are drawn in sharp Vs.

Thread runout

SCHEMATIC REPRESENTATION

External thread

Pitch Crest (thin line)

Root (thick line)

Use alternate long and short lines for representing

crests and roots of the thread, respectively.

Internal thread

SIMPLIFIED REPRESENTATION


Use thick continuous lines for representing crest

and thin continuous lines for representing root of

the thread, respectively.

Pitch/2

Root

Crest

Thread runout

SIMPLIFIED REPRESENTATION


Sectional view

Use thick continuous lines for representing crest

and thin continuous lines for representing root of

the thread, respectively.

ISO (METRIC) THREAD

External

thread

Internal

thread

Center of thread assembly

60o

Pitch, P

P/4

P/8

Thread assemble occurs if and only if both (internal & external)

thread have an equal nominal size (or diameter) and pitch.

Nominal

size

Major

diameter

Pitch Minor

diameter

Tap drill size

M6 6.00 1.00 4.92 5.00

M8 8.00 1.25 6.65 6.75

M10 10.00 1.50 8.38 8.50

M12 12.00 1.75 10.11 10.00

METRIC COARSE THREAD

Minor diameter = Major diameter – Pitch

Minor diameter ≈ Tap drill size

[Table 9.1]

Metric thread

In thread drawing, the following relationship is used.

Nominal

size

Major

diameter

Pitch Minor

diameter

Tap drill size

M8 8.000.75 7.188 7.25

1.00 6.917 7.00

M10 10.00

0.75 9.188 9.25

1.00 8.917 9.00

1.25 8.647 8.75

METRIC FINE THREAD[Table 9.2]

Minor diameter = Major diameter – Pitch

Minor diameter ≈ Tap drill size

In thread drawing, the following relationship is used.

Thread Length

DRAWING STEPS OF

EXTERNAL THREAD

Starting

position

Draw

thread

axis

Ma

jor

dia

.M

ino

r d

ia.

Draw

45o Chamfer

Draw line making

30o with thread axisDraw a circle thatrepresents a crest.

Draw an arc thatrepresents a root.

Internal Thread

1. Through threaded hole

DRAWING STEPS OF

THREADED HOLE

Draw

thread

axis

Ma

jor

dia

.M

ino

r d

ia.

Draw a root

Draw a crestSectional view

Draw a circle thatrepresents a crest.


root

crest

Section lines are drawn into the crest of a thread.

Min

or

dia

.

DRAWING STEPS OF

THREADED HOLE

2. Blinded threaded hole

Draw

thread

axis

Ma

jor

dia

.

Hole depth

Thread depthDraw a circle thatrepresents a crest.


DRAWING STEPS OF

THREADED HOLE

2. Blinded threaded holeSectional view

DIMENSIONING EXTERNAL THREAD

M10 ×1.5

×1.0 Fine thread

Coarse thread

xxThread

length

Use local note to specify :- thread form, nominal size,

pitch (if it is a fine thread)

Use typical method to specify :- thread length.

3. Thread form

4. Nominal size

5. Pitch

1. Tap drill size

2. Drill depth

6. Thread depth

8.50 Drill, 20 Deep,

M10 Tapped, 15 Deep

DIMENSIONING THREADED HOLE

Use local note to

specify

BOLT : Terminology

Bolt is a threaded cylinder with a head.

Hexagonal head

bolt and nutHead

thickness

Thread length

Length

Width

across flat

Dimensions of bolt’s head are

listed in table 9.4.

Draw

bolt

axis

Draw an

end view

of the head

B

Starting

position

H

Draw a

bolt head

B/2 B/2

30o

Length

Thread Length

Ma

jor f

Draw

body of a bolt

BOLT : Drawing steps

NUT : Drawing steps

Draw an end view

of the nut

BH

B/2 B/2

Dimensions of the nut are given in Table 9.14.

Dash lines represent

a threaded hole are

omitted for clarity.

1. Insert a bolt into

a clearance hole

2. Insert a washer.

3. Screw a nut.

BOLT : Application

Let’ s think together…

What do you think about the following suggestions for

design improvement?

(B) Use shorter bolt with the

same thread length.

Correct

(C) Reduce the thread length.

Inappropriate Wrong

(D) Add washer or nut.

(A) Nothing have to be changed.



design improvement?

(B) Use a bolt of this length but

has a longer thread length.

(C) Use a longer bolt with the

same thread length.

(D) Add washer.


Correct Inappropriate Wrong



design improvement ?

(B) Use a bolt of this length but

has a shorter thread length.

(C) Use a longer bolt with the

same thread length.

(D) Add washer.





design improvement?

(A) Use a bolt of this length but

has a shorter thread length.

(B) Use a longer bolt with the

same thread length.

(C) Use a longer bolt by

increasing a thread length

(D) Remove washer.





(A) Increase the bolt diameter.

(B) Use washer with larger

outside diameter.

(C) Reduce the hole diameter.

(D) Add washer at bolt head.





(B) Use a bolt with shorter

thread length.

(C) Add washer.

(D) Increase drill and thread

depths.







(B) Use a bolt with slightly

longer thread length.


Stud is a headless bolt, threaded at both ends.

STUD : Terminology

Thread length

Length

Thread length

Drawing

representation

Ma

jor f

STUD : Drawing steps

Draw

stud

axis Min

or f

Starting

position

Stud Length

Thread Length Thread Length

Draw

45o Chamfer

Draw

45o Chamfer

Draw line making

30o with axis

Chapter 11

Keys and Cotter Joints

Keys

• A key is the piece inserted in an axial direction between a shaft

and hub of the mounted machine element such as pulley or gear

• to prevent relative rotation….

may allow sliding movement along the shaft if required.• Keys are temporary fastening and are always made of mildsteel because they are subjected to shearing and compressivestresses caused by the torque they transmit.• a keyway is the groove cut in the shaft or hub to accommodate akey. Key ways can be milled horizontally or vertically .

Keyways milled

Classification of keys

• Saddle keys» Hollow saddle key

» Flat saddle key

• Sunk keys» Taper sunk keys

» Parallel sunk keys

» Feather keys– Woodruff key (adjustable key)

• Round keys» Parallel pin

» Taper pin

–this key has curved underside so that it can beplaced on the curved surface of the shaft. Thekeyway is cut in the mating piece (hub) only.Hollow saddle key

Saddle key are taper keys and are sunk into the hub onlyTaper prevents axial movement along the shaft

Saddle key is suitable for light duty,

since they rely on a friction drivealone and are liable to slip on the

shaft under load

Flat saddle key

Flat surface onthe shaft

Let D = diameter of the shaftWidth of the key, W = D/4

Nominal thickness at large end, T = W/3 = D/12Length of the key , L = D to1.5DTaper on the top surface =1:100

holding force is comparative largethan the hollow saddle key

Sunk keys

• Sunk keys are sunk in the shaft and the hub. These keysare suitable for heavy duty since they rely on positivedrive.– Taper sunk keys:

• This is the standard form of the key and may be either ofrectangular or square cross-section. The key is sunk in theshaft to a depth of half its nominal thickness when measuredat the side.

» Rectangular cross-section

let D = diameter of the shaft

width of the key W = D/4

nominal thickness T = (2/3)W = (1/6)D

» Square cross-section:

T = W

Sunk taper key

Parallel sunk key

It is uniform in width andthickness throughout.

It is thus taper less and is usedwhere the pulley or othermating piece is required to slidealong the shaft.

It may be rectangular or squarecross-section and their endsmay be squared or rounded.

Feather keys

It is a key attached to one member of a pair

It is a particular kind of parallel key that permits axial moment

A feather key is secured either to the shaft or to the hub, the keybeing a sliding fit in the keyway of the machine element onwhich it moves.

Double HeadedPeg KeyKey

Woodruff keys

The key fits into a semi-circularkeyway in the shaft and the top ofthe key fits into a plain rectangularkey way in the hub of the wheel.

Since the key and the key seat bear thesame radius , it has the advantage ofadjusting itself to any taper of theslot of the hub or boss of wheel

Used in feed gear box of lathe, othermachine tools and in automobiles

Round key or Pin Key:

Joints::: Knuckle joint

Two or more rods subjected to

tensile and compressive forces are

fastened together

Their axes are not in

alignments but meet in a point

The joint allows a

small angular moment

of one rod relative to

another

It can be easily

connected and

disconnected

Applications: Elevator chains, valve rods, etc

Knuckle joint

Cotter joint

Cotter joints with

socket and spigot

ends

for circular bars

Slots are wider than the cotter.

Cotter pulls the rod and socket tightly together

Clearance: must be provided for adjustment.(2 to 3 mm)

Proportionscotter thickness = (1/3)diameter of rod Taper: 1 in 30

Cotter joint

Sleeve and cotter jointFor circular rods

The enlarged ends of the rods butt against each other with a common sleeve over them

•The rod ends are enlarged to take care of the weakening effect caused by slots

Gib and cotter joint for rectangular rods

One bar end is made in the form of a strap

A Gib is used along with the cotter.

Gib is like a cotter but with two gib

heads at its ends .

The thickness of the gib and

cotter are same

Gib and cotter joint or rectangular rods

Chapter 12

Couplings

Couplings

Coupling is a device used to connect two shafts together at their

ends for the purpose of transmitting power

Motor

Coupling

Pump

Uses of coupling

• To provide connection of shafts of units made separately

• To allow misalignment of the shafts or to introduce mechanical flexibility.

• To reduce the transmission of shock loads

• To introduce protection against overloads.

• To alter the vibration characteristics

Types of coupling

• Rigid

• Flexible

• Universal

Rigid coupling

Flexible coupling

Universal coupling

Rigid couplingFlange

Driven Shaft

Driving

Shaft

Key Hub

•Rigid couplings are used when precise shaft alignment is required

•Simple in design and are more rugged

• Generally able to transmit more power than flexible couplings

•Shaft misalignments cannot be compensated

Flanged Coupling

Flexible Coupling

Bush

FlangeFlange

Driving

Shaft

Driven Shaft

Pin

•A flexible coupling permits with in certain limits, relative rotation and

variation in the alignment of shafts

•Pins (Bolts) covered by rubber washer or bush is used connect flanges

with nuts

•The rubber washers or bushes act as a shock absorbers and insulators.

Elastomeric coupling (Tyre Coupling)

Bolt

Flange

Elastomeric member

Flange

•An assembly of components designed to connect axially oriented

shafts in order to provide power transmission

•Able to accommodate shaft misalignment through elastomeric

materials

Advantages and Limitations

Advantages • Torsionally stiff• No lubrication or maintenance• Good vibration damping and shock absorbing qualities• Less expensive than metallic couplings• More misalignment allowable than most metallic couplingsLimitations• Sensitive to chemicals and high temperatures• Usually not torsionally stiff enough for positive displacement• Larger in outside diameter than metallic coupling• Difficult to balance as an assembly

TThank YouANK YOU

Branch: Mechanical Engineering Year: 1st

Documents

Transcript of Branch: Mechanical Engineering Year: 1st